對於一般的全排列問題，使用DFS遞迴方法即可：

n=int(input())
l=[]
for i in range(1,n+1):
    l.append(i)
def swmp(l,i,j):
    l[i],l[j]=l[j],l[i]
def fun(l,k):
    if k==n-1:
        print(l)
        return
    for i in range(k,n):
        swmp(l,k,i)
        fun(l,k+1)
        swmp(l,k,i)
fun(l,0)

但是可以看出，此全排列輸出並不是按字典序排列的：

[1, 2, 3, 4]
[1, 2, 4, 3]
[1, 3, 2, 4]
[1, 3, 4, 2]
[1, 4, 3, 2]
[1, 4, 2, 3]
[2, 1, 3, 4]
[2, 1, 4, 3]
[2, 3, 1, 4]
[2, 3, 4, 1]
[2, 4, 3, 1]
[2, 4, 1, 3]
[3, 2, 1, 4]
[3, 2, 4, 1]
[3, 1, 2, 4]
[3, 1, 4, 2]
[3, 4, 1, 2]
[3, 4, 2, 1]
[4, 2, 3, 1]
[4, 2, 1, 3]
[4, 3, 2, 1]
[4, 3, 1, 2]
[4, 1, 3, 2]
[
 
4, 1, 2, 3

如果要按照字典序排列進行輸出，那麼為了保持大小順序，不只是要調換兩個數的值，而是迴圈移動，程式碼如下：

n=int(input())
l=[]
for i in range(1,n+1):
    l.append(i)
def swmpgo(l,i,j):
    for x in range(j,i,-1):
        l[x-1],l[x]=l[x],l[x-1]
def swmpback(l,i,j):
    for x in range(i,j):
        l[x],l[x+1]=l[x+1],l[x]
def fun(l,k):
    
 
if k==n-1:
        print(l)
        return
    for i in range(k,n):
        swmpgo(l,k,i)
        fun(l,k+1)
        swmpback(l,k,i)
fun(l,0)

此時，全排列的輸出是按照字典順序來的：

[1, 2, 3, 4]
[1, 2, 4, 3]
[1, 3, 2, 4]
[1, 3, 4, 2]
[1, 4, 2, 3]
[1, 4, 3, 2]
[2, 1, 3, 4]
[2, 1, 4, 3]
[2, 3, 1, 4]
[2, 3, 4, 1]
[2, 4, 1, 3]
[2, 4, 3, 1]
[3, 1, 2, 4]
[3, 1, 4, 2]
[3, 2, 1, 4]
[3, 2, 4, 1]
[3, 4, 1, 2]
[3, 4, 2, 1]
[4, 1, 2, 3]
[4, 1, 3, 2]
[4, 2, 1, 3]
[4, 2, 3, 1]
[4, 3, 1, 2]
[4, 3, 2, 1]